# Show that $\sum_{n =1}^\infty a_n$ converges iff $\sum_{k = 1}^\infty ka_{k^2}$ converges.

Let $a_1 \geq a_2 \geq a_3 \geq \cdots \geq 0$. Prove that the series $\sum_{n =1}^\infty a_n$ converges if and only if the series $\sum_{k = 1}^\infty ka_{k^2}$ converges.

I have the following:

Let $S_n = \sum_{k = 1}^na_k$, $t_k = \sum_{j = 1}^k ja_{j^2}$. Let $n \geq k(k+1) / 2)$. Then

\begin{align*} t_k &= a_1 + 2a_4 + 3a_9 + \cdots + ka_{k^2}\\ &\leq a_1 + 2a_3 + 3a_6 + \cdots + ka_{k(k+1)/2}\\ &\leq a_1 + a_2 + a_3 + \cdots + a_{k(k+1)/2}\\ &\leq S_n \end{align*}

So if $\sum_{n =1}^\infty a_n$ converges, then $S_n$ is bounded above, so $t_k$ is bounded above, so $\sum_{k = 1}^\infty ka_{k^2}$ converges since it is a nonnegative series with partial sums bounded above.

How do I prove the other direction?

• $(2k+1)a_{k^2} \geqslant a_{k^2} + a_{k^2+1} + \dotsc + a_{k^2 + 2k}$ – Daniel Fischer Nov 19 '16 at 22:46
• This is a variant of the (Cauchy) condensation test. en.wikipedia.org/wiki/Cauchy_condensation_test, see under "Generalizations". – LutzL Nov 19 '16 at 23:12
• – LutzL Nov 19 '16 at 23:15